问题 解答题
已知向量
.
a
=(m,-1),
.
b
=(
1
2
3
2
),
(Ⅰ)若
a
b
,求实数m的值;
(Ⅱ)若
a
b
,,求实数m的值;
(Ⅲ)若
a
b
,且存在不等于零的实数k,t使得[
a
+(t2-3)
b
]•(-k
a
+t
b
)=0,试求
k+t 2
t
的最小值.
答案

(1)∵

.
a
=(m,-1),
.
b
=(
1
2
3
2
),且
a
b

∴m

3
2
-
1
2
.(-1)=0,∴m=-
3
3

(2)∵

.
a
=(m,-1),
.
b
=(
1
2
3
2
),且
a
b

.
a
.
b
=0,m•
1
2
+(-1)
3
2
=0,∴m=
3

(3)∵

.
a
.
b
,∴
.
a
b
=0.

由条件可得|

a
|=
3
2
+1
 = 2
|b| =
1
2
2
+
3
2
2
=1
,[
a
+(t2-3)
b
]•(-k
a
+t
b
)=0,

即:-k

a
2+(t2-3)t
b
2=0,即-k|
a
|2+(t2-3)t|
b
|2=0,即-4k+(t2-3)t=0.

∴k=

(t2-3)t
4
,由 
k+t2
t
=
t3-3t+4t2
t
=
1
4
(t2+4t-3)=
1
4
(t+2) 2-
7
4

可得当t=-2时,

k+t2
t
有最小值-
7
4

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